Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{-4n + 28}{3n^2 + 6n - 240} \times \dfrac{n^3 + 7n^2 - 30n}{n^2 - 3n} $
Solution: First factor out any common factors. $a = \dfrac{-4(n - 7)}{3(n^2 + 2n - 80)} \times \dfrac{n(n^2 + 7n - 30)}{n(n - 3)} $ Then factor the quadratic expressions. $a = \dfrac {-4(n - 7)} {3(n + 10)(n - 8)} \times \dfrac {n(n + 10)(n - 3)} {n(n - 3)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-4(n - 7) \times n(n + 10)(n - 3) } { 3(n + 10)(n - 8) \times n(n - 3)} $ $a = \dfrac {-4n(n + 10)(n - 3)(n - 7)} {3n(n + 10)(n - 8)(n - 3)} $ Notice that $(n + 10)$ and $(n - 3)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-4n\cancel{(n + 10)}(n - 3)(n - 7)} {3n\cancel{(n + 10)}(n - 8)(n - 3)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $a = \dfrac {-4n\cancel{(n + 10)}\cancel{(n - 3)}(n - 7)} {3n\cancel{(n + 10)}(n - 8)\cancel{(n - 3)}} $ We are dividing by $n - 3$ , so $n - 3 \neq 0$ Therefore, $n \neq 3$ $a = \dfrac {-4n(n - 7)} {3n(n - 8)} $ $ a = \dfrac{-4(n - 7)}{3(n - 8)}; n \neq -10; n \neq 3 $